Physics M.C.Q (1 Marks)

3. i₁ < i₂

Explanation:

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\text{i}=\frac{\text{dQ}}{\text{dt}}$

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\in_0\text{C}\times100\pi\Big[\sin\big(100\pi\text{s}^{-1}\big)\text{t}\Big]\\+\in_0\text{C}\times500\pi\Big[\sin\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$=100\text{C}\pi\in_0\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]\\+500\text{C}\pi\in_0\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big]$

$\text{i}_1=100\pi\in_0\text{C}$ and $\text{i}_2=500\pi\in_0\text{C}$

$\text{i}_2>\text{i}_1$

1. For a given power level, there is a lower current.
2. Lower current implies less power loss.

4. It is easy to reduce the voltage at the receiving end using step-down transformers.

Solution:

Key Concept: Power loss due to transmisssion lines having resistance (R) and I_rms current flowing in the circuit is I²_rms R.

The power is to be transmitted over the large distances ar high alternating voltages, so current flowing through the wires will be low because of given power (P).

For a given power level, we find that

P = E_rmsI_rms (I_rms is low when E_rms is high)

Powor loss = I²_rms R = low ($\because$ I_rms is low)

Now at the receiving end high voltage is reduced by using step-down transformers.

2. $\text{ohm}^{-1}$

Explanation:

Susceptance is the imaginary part of admittance. 

The admittance is the inverse of impedance. 

Unit of admittance is ohm. Unit of admittance is ohm^-1 or siemens. 

Unit of susceptance is same as of admittance.

1. $\frac{10}{2\pi}\text{Hz}$

Explanation:

Resonant frequency $=\frac{1}{2\pi\sqrt{\text{LC}}}$

$=\frac{1}{2\pi\sqrt{10^{-6}}}$

$=\frac{10}{2\pi}\text{Hz}$

4. none of the above is true

Explanation:

The voltages across different elements in AC circuit add vectorially.

The voltages across inductor and capacitor are out of phase and at 90^∘ to that across resistor.

Hence net voltage across source $=\sqrt{(\text{V}_1-\text{V}_2)+\text{V}^2_3}.$

2. $10^4\Omega$

4. $155\Omega$

Explanation:

Inductance of the solenoid $=\text{L}=\frac{\mu\text{N}^2\text{A}}{1}=0.493\text{H}$

reactance of this solenoid $=\text{L}\omega=\text{L}2\pi\text{f}=155\Omega$

1. 50 Hz

Explanation:

$\text{Y}=100\text{V, C}=2\mu\text{F}=20\times10^{-6}$

$\text{I}=0.628\text{A,v}=?$

$\text{X}_\text{C}=\frac{\text{V}}{\text{I}}=\frac{100}{0.628}$

$\Rightarrow\frac{1}{2\pi\text{vC}}=\frac{100}{0.628}$

$\text{v}=\frac{0.628}{100\times2\pi\text{C}}$

$=50\text{ Hz}$

1. 200V

Explanation:

General sinusoidal voltage variation is given by:

$\text{V}=\text{V}_0\sin(\omega\text{t})$

Here, in this equation V₀​ is peak voltage.

So, on comparing both equation we get:

$\text{V}_0=200\text{V},\omega=314$

3. About 310V.

Explanation:

$\text{V}_\text{rms}=220\text{V}$

$\text{V}_\text{p}=\sqrt{2}\times\text{V}_\text{rms}$

$=220\times1.414=311\text{volt}$

3. C, 9C

Explanation:

$\frac{\text{f}_\text{max}}{\text{f}_\text{max}}=3$

$\therefore\frac{\sqrt{\text{LC}_\text{max}}}{\sqrt{\text{LC}_\text{min}}}=3$

$\frac{\text{C}_\text{max}}{\text{C}_\text{min}}=9$

${\text{C}_\text{min}}=\text{C}$

$\therefore{\text{C}_\text{max}}=\text{9C}$

1. Inductor and capacitor.

4. Resistor, inductor and capacitor.

Solution:

Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.

Reactance of an inductor of inductance L is $\text{X}_\text{L}=2\pi\text{vL}$, where v is the frequency of the AC circuit.

X_c = Reactance of the capacitive circuit

$=\frac{1}{2\pi\text{fC}}$

With an increase in frequency in (f) of an AC circuit, R remains constant, inductive reactance (X_L) increases and capacitive reactance (X_C) decreases. For an L-C-R circuit,

Z- Impedance of the circuit,

$=\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}$

$=\sqrt{\text{R}^2+\bigg(2\pi\text{vL}-\frac{1}{2\pi\text{vC}}\bigg)^2}$

As frequency (v) increases, Z decreases and at certain value of the frequency known as resonant frequency (v₀), impedance Z is minimum that is Z_min = R current varies inversely with impedance and at Z_min current is maximum.

4. $10^5\Omega$

Explanation:

$\text{Reactance}=\frac{1}{\text{C}\omega}\Omega$

$=\frac{1}{10^{-6}\times10}=\frac{1}{10^{-5}}$

$=10^5\Omega$

2. 25.6V

Explanation:

Potential difference across the circuit

$\text{V}=\sqrt{\text{V}^2_\text{R}+\text{V}^2_\text{L}}=\sqrt{20^2+16^2}=\sqrt{656}=25.6\text{V}$

1. lags the applied voltage

Explanation:

Capacitive reactance is given by. $\text{X}_\text{C}=\frac{1}{\text{wC}}$

Inductive reactance is given by $\text{X}_\text{L}={\text{wL}}$

At resonance, $\text{X}_\text{L}={\text{X}_\text{C}}\Rightarrow \text{w}\text{L}=\frac{1}{\text{wC}}$

But a frequency higher than resonance frequency, X_L​ > X_C​

So the circuit behaves as a inductive circuit at a frequency higher that resonant frequency and the current lags behind the voltage in an inductive circuit.

3. 90°

Explanation:

$\text{i}=\text{i}_\text{o}\sin(\omega\text{t}+\frac{\pi}{2})$

current leads voltage by $\frac{\pi}{2}$ i.e., current and voltage differ in phase by 90°

2. May be 1000W.

4. May be less than 1000W.

Explanation:

Average power $\text{P}_\text{av}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$

$=100\times10\cos\phi$

$\text{P}_\text{av}=1000\cos\phi$

$\therefore\ \cos\phi$ lies "0 to 1".

$\Rightarrow\ 0\leq\text{P}_\text{av}\leq1000.$

1. 50V

Explanation:

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{L}-\text{V}_\text{C})^2}=\sqrt{30^2+(60-20)^2}=50\text{V}$

3. Voltage is more but current is less

Explanation:

The power cables have some resistance. 

Power lost in the wires can be calculated as P = I²R with R as the resistance of the wires and I as the current that passes through them.

Power at the load is P = VI. 

From this one can see that if voltage is increased by say n times, then only $\frac{1}{\text{n}}$​ the current is required to deliver the same power. However, if $\frac{1}{\text{n}}$​ current is passed

on the same wires, only $\frac{1}{\text{n}^2}$ of the power will be lost.

1. An inductor and a capacitor.

Explanation:

$\text{X}=0$ (Given)

$\text{X}=\text{X}_\text{L}+\text{X}_\text{C}$

$=\omega\text{L}-\frac{1}{\omega\text{C}}=0$

It is possible that the circuit contains an inductor and a capacitor.

1. Of the inductor increases.

Explanation:

$\text{X}_\text{L}=\omega\text{L}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$

If frequency increases that causes 'X_L' raction of inductor increases and 'X_C' reactance of capacitor decreses.

2. $\frac{\pi}{2}$

Explanation:

Current leads voltage by $\frac{\pi}{2}$

$\therefore$ phase difference is $=\frac{\pi}{2}$

4. (d)

Explanation:

$\text{X}=\text{X}_\text{L}-\text{X}_\text{C}$

$=\omega\text{L}-\frac{1}{\omega\text{C}}$

$=2\pi\text{fL}-\frac{1}{2\pi\text{fC}}$

3. 311V

Explanation:

The rated voltage in bulb is rms voltage.

$\text{V}_\text{rms}=\frac{\text{V}_0}{\sqrt{2}}$

$\text{V}_0=\sqrt{2}\times220=311.08\text{V}$

2. $\big(\frac{10}{\sqrt{2\text{V}}}\big)$

4. About 10A.

Explanation:

$\text{I}_\text{p}=14\text{Amp}$

$\text{I}_\text{rms}=\frac{\text{IP}}{\sqrt{2}}=\frac{14}{\sqrt{2}}$

$=9.9=10\text{Amp}.$

2. Capacitive reactance is inversely proportional to the frequency of the current

Explanation:

Capacitative reactance is an opposition to the change of voltage across an element.

It is denoted by XC​ and is inversely proportional to the signal frequency (f) and the capacitance C

$\text{X}_\text{c}=\frac{1}{2\pi\text{fc}}$

1. $2\text{C}$

Explanation:

Resonance frequency, $(\text{f})=\frac{1}{2\pi\sqrt{\text{LC}}}$

For f to be constant the product LC must be constant.

So, if we half the value of inductance then the value of capacitance must be doubled.

C should be changed to 2C.

2. $157\Omega$

Explanation:

$\text{Reactance }=\omega\text{L}$

$=2\pi\text{fL}$

$=2\pi\times50\times0.5$

$157\Omega$

3. Admittance

Explanation:

Impedance is the opposition a circuit presents to a current when a voltage is applied.

Admittance is a measure of how easily a circuit or device will allow a current to flow.

Admittance is defined as $\text{Y}=\frac{1}{\text{Z}}$

where Z is the impedance of the circuit.

4. Zero

Explanation:

Power, $\text{P}=\text{I}_\text{rms}\times\text{V}_\text{rms}\times\cos\phi$

In the given problem, the phase difference between voltage and current is $\frac{\text{P}}{2}$ Hence

$\text{P}=\text{I}_\text{rms}\times\text{V}_\text{rms}\times\cos\big(\frac{\pi}{2}\big)=0$

4. None of the above.

Explanation:

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{C}-\text{V}_\text{L})^2}=10\text{V}$

V_C​ > V_L​, hence current leads the voltage.

Power factor $=\cos\phi=\frac{8}{10}=0.8$

3. LC will have to be decreased

Explanation:

$\text{Resonant frequency}=\frac{1}{\sqrt{\text{LC}}}$

$\text{LC}\downarrow\text{if }\omega_\text{r}\uparrow$

2. $\text{R}$

Explanation:

At resonance $\frac{1}{\omega\text{C}}=\omega\text{L}$

$\therefore$ impedance = R

3. $30\pi$

1. Decrease to one-half of the original value

Explanation:

Resonant frequency in series LCR circuit: $\omega=\sqrt{\frac{1}{\text{LC}}}$

Resonant frequency in series LCR circuit, $\omega=\sqrt{\frac{1}{\text{LC}}}=\sqrt{\frac{1}{\text{2L}\times\text{2C}}}=\frac{\omega}{2}$

4. lags behind the voltage by $\frac{\pi}{2}$

Explanation:

In a purely inductive circuit (an AC circuit containing inductance only) the current lags behind the voltage by $\frac{\pi}{2}$

1. DC.

Explanation:

$\text{X}_\text{C}\frac{1}{\omega\text{C}}=\frac{1}{0\times\text{C}}$ $\bigg\{\text{in}\stackrel{{\text{DC}}}{{\omega = 0 }}\bigg\}$

$=\infty$